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A frictionless wire is fixed between A ...

A frictionless wire is fixed between `A` and `B` inside of a hollow sphere of radius `R` as shown. A beam slips along the wire starting from the rest at point `A`. The time taken by the beam to slip from `A` to `B` will be

A

`2sqrt(R//g)`

B

`gRsqrt(gcostheta)`

C

`(2sqrt(gR))/(gcostheta)`

D

`(2sqrt(gRcostheta))/(g)`

Text Solution

Verified by Experts

The correct Answer is:
A
`AB=2R cos theta`
acceleration along `AB`
`a=g cos theta`
`u=0` from `A` to `B`
`S=ut+(1)/(2)at^(2)`

`2R cos theta=0+(1)/(2)(g cos theta)t^(2)`
`t=2sqrt((R)/(g))`
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